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Solve this

Asked by venkat 25th December 2017, 9:03 PM
Answered by Expert
Answer:
Time taken for the lead ball to reach top surface of lake is given by
begin mathsize 12px style h space equals space u cross times t space plus space 1 half cross times g cross times t squared
sin c e space i n i t i a l space v e l o c i t y space u equals 0 comma space space h space equals space 1 half cross times g cross times t squared
h e n c e space t space equals space square root of fraction numerator 2 cross times g over denominator h end fraction end root space equals space square root of fraction numerator 2 cross times 10 over denominator 5 end fraction end root space equals space 2 s end style
 
velocity of the ball when it reach the lake surface is given by
 
begin mathsize 12px style v space equals space u space plus space g cross times t

sin c e space i n i t i a l space v e l o c i t y space u equals 0. space v space equals space g cross times t space equals space 10 cross times 2 space equals space 20 space m divided by s
end style
For another 1 second, lead ball travelled with velocity 20 m/s. hence the distance travelled, depth of lake is 20 m
Answered by Expert 27th December 2017, 2:14 PM
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